∵AD, DC, BG intercept ⊙ O at point A, point E and point B respectively.

∴DA=DE,CE=CB，

Let BC be X, then CF=x-2 and DC=x+2.

In Rt△DFC, (x+2)2-(x-2)2=(2 5)2,

Solution: x = 5/2;

∫AD∨BG，

∴∠DAE=∠EGC，

DA = DE，

∴∠dae=∠aed；

∠∠AED =∠CEG，

∴∠EGC=∠CEG，

∴CG=CE=CB= 5/2，

∴BG=5，

∴AG= (2 root number 5) 2+5 2 = root number 45=3 root number 5;

Solution 1: connect BE, S△ABG= 1/2AB? BG= 1/2AG？ Yes,

∴ 2 5×5=3 5BE,

∴ BE= 10/3，

In Rt△BEG,

EG= radical number (BG 2-be 2) = radical number [5 2-( 10/3) 2] = 5/3 radical number 5,

Solution 2: 2:∠∠DAE =∠EGC, ∠AED=∠CEG,

∴△ADE∽△GCE，

∴ AD/CG=AE/EG, 5/2.5=3 radical 5-EG/EG, the solution is EG=5 radical 5/3.

Some 2' s are square, you can understand.